Kaylamv2007
Kaylamv2007 Kaylamv2007
  • 10-03-2020
  • Mathematics
contestada

How do you solve 4x^2 - 8x + 4 = 0 as completing the square? Is there a solution?

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example193 example193
  • 10-03-2020

Answer:

1

Step-by-step explanation:

Yes, there is a solution! You first look for a common factor to factor away. Observing, you see that there is common multiple of 4, so your equation becomes [tex]4(x^2-2x+1)=0[/tex]. Dividing away the 4, we  get [tex]x^2-2x+1=0[/tex]. From here, we can complete the square: [tex](x-\frac{2}{2})^2-\frac{2}{2}+1=0[/tex] which is equal to [tex](x-1)^2=0[/tex]. When x=1, both sides of the equation equal 0, so x=1.

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